Water ionizes by the equation H2O(l)?H+(aq)+OH?(aq) The extent of the reaction i

Question

Water ionizes by the equation H2O(l)?H+(aq)+OH?(aq) The extent of the reaction is small in pure water and dilute aqueous solutions. This reaction creates the following relationship between [H+] and [OH?]: Kw=[H+][OH?] Keep in mind that, like all equilibrium constants, the value of Kw changes with temperature. Part A - What is the H+ concentration for an aqueous solution with pOH = 4.06 at 25 ?C? At a certain temperature, the pH of a neutral solution is 7.34. Part B - What is the value of Kw at that temperature

Explanation / Answer

1)pH = 14 - pOH
pH = 9.94
H+ concen = 10^-pH
H+ = 1.14x 10^ -10M

2)It is very important to know that neutral pH changes with temperature. The key part to this problem is that the solution is neutral! In a neutral solution the pH and pOH are equal and therefor so are their concentrations.

Kw= [H]*[OH] where [H] and [OH] represent the concentrations of those ions.

pH=-log[H]
antilog(-pH)=[H]

Antilog is fancy way of saying 10^x. In our case x=-7.34

[H]=4.57 x 10^-8 M (Molarity)

Since pH and pOH are equal in our neutral solution so are their concentrations.

[OH]=4.57 x 10^-8 M

Now we go back to Kw= [H]*[OH] and plug in the concentrations.

Kw=(2.95 x 10^-8)*(2.95 x 10^-8)=20.88 x 10^-16

This method works for any temperature not just 25 degrees Celsius.

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